Question Number 119421 by bemath last updated on 24/Oct/20
$${find}\:{the}\:{value}\:{of}\:\underset{{i}=\mathrm{1}} {\overset{\mathrm{999}} {\prod}}\:\mathrm{cos}\:\left({ia}\right)\:;\:{where}\:{a}\:=\:\frac{\mathrm{2}\pi}{\mathrm{1999}} \\ $$
Commented by MJS_new last updated on 24/Oct/20
$$\mathrm{study}\:{P}_{{n}} =\underset{{j}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cos}\:\frac{\mathrm{2}\pi{j}}{\mathrm{2}{n}+\mathrm{1}}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{answer} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{easier}\:\mathrm{than}\:\mathrm{it}\:\mathrm{seems} \\ $$
Answered by benjo_mathlover last updated on 24/Oct/20
$${Let}\:{B}\:{denote}\:{the}\:{desired}\:{product},\:{and}\:{let} \\ $$$${J}\:=\:\mathrm{sin}\:{a}\:\mathrm{sin}\:\mathrm{2}{a}\:\mathrm{sin}\:\mathrm{3}{a}…\:\mathrm{sin}\:\mathrm{999}{a} \\ $$$${consider}\:{that}\: \\ $$$$\mathrm{2}^{\mathrm{999}} \:{BJ}\:=\:\left(\mathrm{2sin}\:{a}\:\mathrm{cos}\:{a}\right)\left(\mathrm{2sin}\:\mathrm{2}{a}\mathrm{cos}\:\mathrm{2}{a}\right)\left(\mathrm{2sin}\:\mathrm{3}{a}\mathrm{cos}\:\mathrm{3}{a}\right)…\left(\mathrm{2sin}\:\mathrm{999}{a}\:\mathrm{cos}\:\mathrm{999}{a}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{sin}\:\mathrm{2}{a}\:\mathrm{sin}\:\mathrm{4}{a}\:\mathrm{sin}\:\mathrm{6}{a}\:…\:\mathrm{sin}\:\mathrm{1998}{a} \\ $$$$\:\:\:\:\:\:\:=\:\left(\mathrm{sin}\:\mathrm{2}{a}\mathrm{sin}\:\mathrm{4}{a}\mathrm{sin}\:\mathrm{6}{a}…\mathrm{sin}\:\mathrm{998}{a}\right)\left(−\mathrm{sin}\:\left(\mathrm{2}\pi−\mathrm{1000}{a}\right)\right).\left(−\mathrm{sin}\:\left(\mathrm{2}\pi−\mathrm{10002}{a}\right)\right)…\left(−\mathrm{sin}\:\left(\mathrm{2}\pi−\mathrm{1998}{a}\right)\right) \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{sin}\:\mathrm{2}{a}\mathrm{sin}\:\mathrm{4}{a}\mathrm{sin}\:\mathrm{6}{a}…\mathrm{sin}\:\mathrm{998}{a}\mathrm{sin}\:\mathrm{999}{a}\mathrm{sin}\:\mathrm{997}{a}…\mathrm{sin}\:{a}\:=\:{J} \\ $$$${since}\:{J}\:\neq\:\mathrm{0}\:,\:{hence}\:{the}\:{desired}\:{product} \\ $$$${is}\:{B}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{999}} }\: \\ $$
Answered by mindispower last updated on 24/Oct/20
![Π_(i=1) ^m cos(i((2π)/(2m+1))) Z^(2m+1) −1=0⇒Z=e^((2ikπ)/(2m+1)) ,k∈[0,2m] z^(2m+1) −1=Π_(k≤2m) (z−e^((2ikπ)/(2m+1)) ) z=−1⇒ −2=Π_(k=0) ^(2m) (−1−e^((2ikπ)/(2m+1)) )=−1e^(i(π/(2m+1)).((2m(2m+1))/2)) .Π_(k≤2m) (2cos(((kπ)/(2m+1))) =(−1)^(m+1) 2^(2m+1) Π_(k≤2m) cos(((kπ)/(2m+1))) Π_(k≤2m) cos(((kπ)/(2m+1)))=Π_(k≤m) cos(((kπ)/(2m+1))).Π_(m<k≤2m) (−cos(π−((kπ)/(2m+1)))) put k→2m+1−k in2nd ⇒ =Π_(1≤k≤m) cos(((kπ)/(2m+1))).(−1)^m .Π_(k≤m) cos(((π(2m+1)−(2m+1−k)π)/(2m+1)) =(−1)^m (Π_(k=1) ^m cos(((kπ)/(2m+1))))^2 ⇒−2=(−1)^(m+1) .2^(2m+1) .(−1)^m .(Π_(k≤m) cos(((kπ)/(2m+1))))^2 ⇒(1/2^(2m) )=Π_(k≤m) cos^2 (((kπ)/(2m+1))) since0<((kπ)/(2m+1))<((mπ)/(2m))=(π/2) ⇒Π_(k≤m) cos(((kπ)/(2m+1)))=(1/2^m ) put m=999⇒Π_(k≤999) cos(((kπ)/(1999)))=(1/2^(999) )](https://www.tinkutara.com/question/Q119443.png)
$$\underset{{i}=\mathrm{1}} {\overset{{m}} {\prod}}{cos}\left({i}\frac{\mathrm{2}\pi}{\mathrm{2}{m}+\mathrm{1}}\right) \\ $$$${Z}^{\mathrm{2}{m}+\mathrm{1}} −\mathrm{1}=\mathrm{0}\Rightarrow{Z}={e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{2}{m}+\mathrm{1}}} ,{k}\in\left[\mathrm{0},\mathrm{2}{m}\right] \\ $$$${z}^{\mathrm{2}{m}+\mathrm{1}} −\mathrm{1}=\underset{{k}\leqslant\mathrm{2}{m}} {\prod}\left({z}−{e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{2}{m}+\mathrm{1}}} \right) \\ $$$${z}=−\mathrm{1}\Rightarrow \\ $$$$−\mathrm{2}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{m}} {\prod}}\left(−\mathrm{1}−{e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{2}{m}+\mathrm{1}}} \right)=−\mathrm{1}{e}^{{i}\frac{\pi}{\mathrm{2}{m}+\mathrm{1}}.\frac{\mathrm{2}{m}\left(\mathrm{2}{m}+\mathrm{1}\right)}{\mathrm{2}}} .\underset{{k}\leqslant\mathrm{2}{m}} {\prod}\left(\mathrm{2}{cos}\left(\frac{{k}\pi}{\mathrm{2}{m}+\mathrm{1}}\right)\right. \\ $$$$=\left(−\mathrm{1}\right)^{{m}+\mathrm{1}} \mathrm{2}^{\mathrm{2}{m}+\mathrm{1}} \underset{{k}\leqslant\mathrm{2}{m}} {\prod}{cos}\left(\frac{{k}\pi}{\mathrm{2}{m}+\mathrm{1}}\right) \\ $$$$\underset{{k}\leqslant\mathrm{2}{m}} {\prod}{cos}\left(\frac{{k}\pi}{\mathrm{2}{m}+\mathrm{1}}\right)=\underset{{k}\leqslant{m}} {\prod}{cos}\left(\frac{{k}\pi}{\mathrm{2}{m}+\mathrm{1}}\right).\underset{{m}<{k}\leqslant\mathrm{2}{m}} {\prod}\left(−{cos}\left(\pi−\frac{{k}\pi}{\mathrm{2}{m}+\mathrm{1}}\right)\right) \\ $$$${put}\:{k}\rightarrow\mathrm{2}{m}+\mathrm{1}−{k}\:{in}\mathrm{2}{nd}\:\Rightarrow \\ $$$$=\underset{\mathrm{1}\leqslant{k}\leqslant{m}} {\prod}{cos}\left(\frac{{k}\pi}{\mathrm{2}{m}+\mathrm{1}}\right).\left(−\mathrm{1}\right)^{{m}} .\underset{{k}\leqslant{m}} {\prod}{cos}\left(\frac{\pi\left(\mathrm{2}{m}+\mathrm{1}\right)−\left(\mathrm{2}{m}+\mathrm{1}−{k}\right)\pi}{\mathrm{2}{m}+\mathrm{1}}\right. \\ $$$$=\left(−\mathrm{1}\right)^{{m}} \left(\underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}{cos}\left(\frac{{k}\pi}{\mathrm{2}{m}+\mathrm{1}}\right)\right)^{\mathrm{2}} \\ $$$$\Rightarrow−\mathrm{2}=\left(−\mathrm{1}\right)^{{m}+\mathrm{1}} .\mathrm{2}^{\mathrm{2}{m}+\mathrm{1}} .\left(−\mathrm{1}\right)^{{m}} .\left(\underset{{k}\leqslant{m}} {\prod}{cos}\left(\frac{{k}\pi}{\mathrm{2}{m}+\mathrm{1}}\right)\right)^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{m}} }=\underset{{k}\leqslant{m}} {\prod}{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{m}+\mathrm{1}}\right) \\ $$$${since}\mathrm{0}<\frac{{k}\pi}{\mathrm{2}{m}+\mathrm{1}}<\frac{{m}\pi}{\mathrm{2}{m}}=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\underset{{k}\leqslant{m}} {\prod}{cos}\left(\frac{{k}\pi}{\mathrm{2}{m}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{2}^{{m}} } \\ $$$${put}\:{m}=\mathrm{999}\Rightarrow\underset{{k}\leqslant\mathrm{999}} {\prod}{cos}\left(\frac{{k}\pi}{\mathrm{1999}}\right)=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{999}} } \\ $$$$ \\ $$